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Prime number raised to a power


question on prime numberPrime number questionFind the prime-power decomposition of 999999999999If all pairs of addends that sum up to $N$ are coprime, then $N$ is prime.Is the square of a prime necessarily larger than the next prime?There are infinitely many odd numbers not expressible as the sum of a prime number and a power of $2$Is there any deep reason that 23456789 is prime?Is this number prime?Advise on determining prime numbers such that the product of their geometric series is equivalent to another prime number raised to a power.Is there a name for a multiplicative function $psi(p^n) = (p^n-1)(p^n-1-1)cdots(p-1)$?






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








2












$begingroup$


Let $p$ be a prime number. If $n, a$ are two positive integers such that $p^n | a$, then $p|a$.

I've recently come across this result in a problem. They didn't give any proof for it, they just asserted that it is true. I am not too good at number theory, so please tell me whether this is some well-known result.






number-theory elementary-number-theory







share|cite|improve this question









$endgroup$







  • 2




    $begingroup$
    $p^n|aiff a=p^nmiff dfrac ap=p^n-1minmathbb N$
    $endgroup$
    – Piquito
    8 hours ago

















2












$begingroup$


Let $p$ be a prime number. If $n, a$ are two positive integers such that $p^n | a$, then $p|a$.

I've recently come across this result in a problem. They didn't give any proof for it, they just asserted that it is true. I am not too good at number theory, so please tell me whether this is some well-known result.






number-theory elementary-number-theory







share|cite|improve this question









$endgroup$







  • 2




    $begingroup$
    $p^n|aiff a=p^nmiff dfrac ap=p^n-1minmathbb N$
    $endgroup$
    – Piquito
    8 hours ago













2












2








2


1



$begingroup$


Let $p$ be a prime number. If $n, a$ are two positive integers such that $p^n | a$, then $p|a$.

I've recently come across this result in a problem. They didn't give any proof for it, they just asserted that it is true. I am not too good at number theory, so please tell me whether this is some well-known result.






number-theory elementary-number-theory







share|cite|improve this question









$endgroup$




Let $p$ be a prime number. If $n, a$ are two positive integers such that $p^n | a$, then $p|a$.

I've recently come across this result in a problem. They didn't give any proof for it, they just asserted that it is true. I am not too good at number theory, so please tell me whether this is some well-known result.






number-theory elementary-number-theory




number-theory elementary-number-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 8 hours ago









Math GuyMath Guy

1807 bronze badges




1807 bronze badges







  • 2




    $begingroup$
    $p^n|aiff a=p^nmiff dfrac ap=p^n-1minmathbb N$
    $endgroup$
    – Piquito
    8 hours ago












  • 2




    $begingroup$
    $p^n|aiff a=p^nmiff dfrac ap=p^n-1minmathbb N$
    $endgroup$
    – Piquito
    8 hours ago







2




2




$begingroup$
$p^n|aiff a=p^nmiff dfrac ap=p^n-1minmathbb N$
$endgroup$
– Piquito
8 hours ago




$begingroup$
$p^n|aiff a=p^nmiff dfrac ap=p^n-1minmathbb N$
$endgroup$
– Piquito
8 hours ago










5 Answers
5






active

oldest

votes


















4












$begingroup$

We have $p^nmid a$ and obviously $pmid p^n$. Then $pmid a$ by transitivity.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Thank you! Which part uses specifically that $p$ is a prime number?
    $endgroup$
    – Math Guy
    8 hours ago






  • 2




    $begingroup$
    This does not depend on $p$ prime. It works for any $n$.
    $endgroup$
    – Antonios-Alexandros Robotis
    8 hours ago






  • 2




    $begingroup$
    Nowhere, we do not need prime numbers here.
    $endgroup$
    – Azif00
    8 hours ago


















4












$begingroup$

We have that $p mid p^n$ and that $p^n mid a$, such that $p mid a$. Maybe try to prove for yourself that divisibility is transitive.



Otherwise, let me prove it for you here (at least for you to compare):



Suppose that $a mid b$ and $b mid c$, that means we have $aa' = b$ and $bb' = c$ for some integers $a'$ and $b'$. We get $c = bb' = (aa')b' = a(a'b')$, such that $a mid c$.






share|cite|improve this answer









$endgroup$




















    3












    $begingroup$

    If $n = 1$, we have $p mid a$ and we are done.



    For $n ge 2$:



    $p^n mid a Longrightarrow exists c, ; a = cp^n = (cp^n - 1)p Longrightarrow p mid a. tag 1$



    This result, being so simple, is very well-known.



    We note that $p$ needn't be prime.






    share|cite|improve this answer











    $endgroup$




















      2












      $begingroup$

      Maybe this is an easier way to look at this:
      For integers $a,b,c$, if $b$ divides $c$, and $a$ divides $b$, then $a$ divides $c$.






      share|cite|improve this answer









      $endgroup$




















        0












        $begingroup$

        If $p^n | a$ then $a = p^nk$ for some integer $k$.



        As $p | p^kk$, so we have $p|a$.






        share|cite|improve this answer









        $endgroup$















          Your Answer








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          5 Answers
          5






          active

          oldest

          votes








          5 Answers
          5






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          4












          $begingroup$

          We have $p^nmid a$ and obviously $pmid p^n$. Then $pmid a$ by transitivity.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            Thank you! Which part uses specifically that $p$ is a prime number?
            $endgroup$
            – Math Guy
            8 hours ago






          • 2




            $begingroup$
            This does not depend on $p$ prime. It works for any $n$.
            $endgroup$
            – Antonios-Alexandros Robotis
            8 hours ago






          • 2




            $begingroup$
            Nowhere, we do not need prime numbers here.
            $endgroup$
            – Azif00
            8 hours ago















          4












          $begingroup$

          We have $p^nmid a$ and obviously $pmid p^n$. Then $pmid a$ by transitivity.






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            Thank you! Which part uses specifically that $p$ is a prime number?
            $endgroup$
            – Math Guy
            8 hours ago






          • 2




            $begingroup$
            This does not depend on $p$ prime. It works for any $n$.
            $endgroup$
            – Antonios-Alexandros Robotis
            8 hours ago






          • 2




            $begingroup$
            Nowhere, we do not need prime numbers here.
            $endgroup$
            – Azif00
            8 hours ago













          4












          4








          4





          $begingroup$

          We have $p^nmid a$ and obviously $pmid p^n$. Then $pmid a$ by transitivity.






          share|cite|improve this answer









          $endgroup$



          We have $p^nmid a$ and obviously $pmid p^n$. Then $pmid a$ by transitivity.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 8 hours ago









          Azif00Azif00

          9479 bronze badges




          9479 bronze badges











          • $begingroup$
            Thank you! Which part uses specifically that $p$ is a prime number?
            $endgroup$
            – Math Guy
            8 hours ago






          • 2




            $begingroup$
            This does not depend on $p$ prime. It works for any $n$.
            $endgroup$
            – Antonios-Alexandros Robotis
            8 hours ago






          • 2




            $begingroup$
            Nowhere, we do not need prime numbers here.
            $endgroup$
            – Azif00
            8 hours ago
















          • $begingroup$
            Thank you! Which part uses specifically that $p$ is a prime number?
            $endgroup$
            – Math Guy
            8 hours ago






          • 2




            $begingroup$
            This does not depend on $p$ prime. It works for any $n$.
            $endgroup$
            – Antonios-Alexandros Robotis
            8 hours ago






          • 2




            $begingroup$
            Nowhere, we do not need prime numbers here.
            $endgroup$
            – Azif00
            8 hours ago















          $begingroup$
          Thank you! Which part uses specifically that $p$ is a prime number?
          $endgroup$
          – Math Guy
          8 hours ago




          $begingroup$
          Thank you! Which part uses specifically that $p$ is a prime number?
          $endgroup$
          – Math Guy
          8 hours ago




          2




          2




          $begingroup$
          This does not depend on $p$ prime. It works for any $n$.
          $endgroup$
          – Antonios-Alexandros Robotis
          8 hours ago




          $begingroup$
          This does not depend on $p$ prime. It works for any $n$.
          $endgroup$
          – Antonios-Alexandros Robotis
          8 hours ago




          2




          2




          $begingroup$
          Nowhere, we do not need prime numbers here.
          $endgroup$
          – Azif00
          8 hours ago




          $begingroup$
          Nowhere, we do not need prime numbers here.
          $endgroup$
          – Azif00
          8 hours ago













          4












          $begingroup$

          We have that $p mid p^n$ and that $p^n mid a$, such that $p mid a$. Maybe try to prove for yourself that divisibility is transitive.



          Otherwise, let me prove it for you here (at least for you to compare):



          Suppose that $a mid b$ and $b mid c$, that means we have $aa' = b$ and $bb' = c$ for some integers $a'$ and $b'$. We get $c = bb' = (aa')b' = a(a'b')$, such that $a mid c$.






          share|cite|improve this answer









          $endgroup$

















            4












            $begingroup$

            We have that $p mid p^n$ and that $p^n mid a$, such that $p mid a$. Maybe try to prove for yourself that divisibility is transitive.



            Otherwise, let me prove it for you here (at least for you to compare):



            Suppose that $a mid b$ and $b mid c$, that means we have $aa' = b$ and $bb' = c$ for some integers $a'$ and $b'$. We get $c = bb' = (aa')b' = a(a'b')$, such that $a mid c$.






            share|cite|improve this answer









            $endgroup$















              4












              4








              4





              $begingroup$

              We have that $p mid p^n$ and that $p^n mid a$, such that $p mid a$. Maybe try to prove for yourself that divisibility is transitive.



              Otherwise, let me prove it for you here (at least for you to compare):



              Suppose that $a mid b$ and $b mid c$, that means we have $aa' = b$ and $bb' = c$ for some integers $a'$ and $b'$. We get $c = bb' = (aa')b' = a(a'b')$, such that $a mid c$.






              share|cite|improve this answer









              $endgroup$



              We have that $p mid p^n$ and that $p^n mid a$, such that $p mid a$. Maybe try to prove for yourself that divisibility is transitive.



              Otherwise, let me prove it for you here (at least for you to compare):



              Suppose that $a mid b$ and $b mid c$, that means we have $aa' = b$ and $bb' = c$ for some integers $a'$ and $b'$. We get $c = bb' = (aa')b' = a(a'b')$, such that $a mid c$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered 8 hours ago









              ThorWittichThorWittich

              1,8061 silver badge12 bronze badges




              1,8061 silver badge12 bronze badges





















                  3












                  $begingroup$

                  If $n = 1$, we have $p mid a$ and we are done.



                  For $n ge 2$:



                  $p^n mid a Longrightarrow exists c, ; a = cp^n = (cp^n - 1)p Longrightarrow p mid a. tag 1$



                  This result, being so simple, is very well-known.



                  We note that $p$ needn't be prime.






                  share|cite|improve this answer











                  $endgroup$

















                    3












                    $begingroup$

                    If $n = 1$, we have $p mid a$ and we are done.



                    For $n ge 2$:



                    $p^n mid a Longrightarrow exists c, ; a = cp^n = (cp^n - 1)p Longrightarrow p mid a. tag 1$



                    This result, being so simple, is very well-known.



                    We note that $p$ needn't be prime.






                    share|cite|improve this answer











                    $endgroup$















                      3












                      3








                      3





                      $begingroup$

                      If $n = 1$, we have $p mid a$ and we are done.



                      For $n ge 2$:



                      $p^n mid a Longrightarrow exists c, ; a = cp^n = (cp^n - 1)p Longrightarrow p mid a. tag 1$



                      This result, being so simple, is very well-known.



                      We note that $p$ needn't be prime.






                      share|cite|improve this answer











                      $endgroup$



                      If $n = 1$, we have $p mid a$ and we are done.



                      For $n ge 2$:



                      $p^n mid a Longrightarrow exists c, ; a = cp^n = (cp^n - 1)p Longrightarrow p mid a. tag 1$



                      This result, being so simple, is very well-known.



                      We note that $p$ needn't be prime.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited 8 hours ago

























                      answered 8 hours ago









                      Robert LewisRobert Lewis

                      50.8k2 gold badges33 silver badges69 bronze badges




                      50.8k2 gold badges33 silver badges69 bronze badges





















                          2












                          $begingroup$

                          Maybe this is an easier way to look at this:
                          For integers $a,b,c$, if $b$ divides $c$, and $a$ divides $b$, then $a$ divides $c$.






                          share|cite|improve this answer









                          $endgroup$

















                            2












                            $begingroup$

                            Maybe this is an easier way to look at this:
                            For integers $a,b,c$, if $b$ divides $c$, and $a$ divides $b$, then $a$ divides $c$.






                            share|cite|improve this answer









                            $endgroup$















                              2












                              2








                              2





                              $begingroup$

                              Maybe this is an easier way to look at this:
                              For integers $a,b,c$, if $b$ divides $c$, and $a$ divides $b$, then $a$ divides $c$.






                              share|cite|improve this answer









                              $endgroup$



                              Maybe this is an easier way to look at this:
                              For integers $a,b,c$, if $b$ divides $c$, and $a$ divides $b$, then $a$ divides $c$.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered 8 hours ago









                              Ruben du BurckRuben du Burck

                              9332 silver badges14 bronze badges




                              9332 silver badges14 bronze badges





















                                  0












                                  $begingroup$

                                  If $p^n | a$ then $a = p^nk$ for some integer $k$.



                                  As $p | p^kk$, so we have $p|a$.






                                  share|cite|improve this answer









                                  $endgroup$

















                                    0












                                    $begingroup$

                                    If $p^n | a$ then $a = p^nk$ for some integer $k$.



                                    As $p | p^kk$, so we have $p|a$.






                                    share|cite|improve this answer









                                    $endgroup$















                                      0












                                      0








                                      0





                                      $begingroup$

                                      If $p^n | a$ then $a = p^nk$ for some integer $k$.



                                      As $p | p^kk$, so we have $p|a$.






                                      share|cite|improve this answer









                                      $endgroup$



                                      If $p^n | a$ then $a = p^nk$ for some integer $k$.



                                      As $p | p^kk$, so we have $p|a$.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered 7 hours ago









                                      mlchristiansmlchristians

                                      44414 bronze badges




                                      44414 bronze badges



























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                                          Ласкавець круглолистий Зміст Опис | Поширення | Галерея | Примітки | Посилання | Навігаційне меню58171138361-22960890446Bupleurum rotundifoliumEuro+Med PlantbasePlants of the World Online — Kew ScienceGermplasm Resources Information Network (GRIN)Ласкавецькн. VI : Літери Ком — Левиправивши або дописавши її

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                                          ЛаскавецьФлора УкраїниФлора ЄвропиФлора АфрикиФлора АзіїРослини, описані 1753 ОкружковіОднорічнатрав'яниста рослинаАфриціЄвропіАзіїУкраїні Ласкавець круглолистий Матеріал з Вікіпедії — вільної енциклопедії. Перейти до навігації Перейти до пошуку ? Ласкавець круглолистий Біологічна класифікація Домен: Ядерні (Eukaryota) Царство: Зелені рослини (Viridiplantae) Відділ: Вищі рослини (Streptophyta) — Судинні (Tracheophyta) — Покритонасінні (Angiospermae) — Евдикоти (Eudicots) — Айстериди (Asterids) Порядок: Аралієцвіті (Apiales) Родина: Окружкові (Apiaceae) Підродина: Селерові (Apioideae) Триба: Bupleureae Рід: Ласкавець ( Bupleurum ) Вид: Ласкавець круглолистий Біноміальна назва Bupleurum rotundifolium L., 1753 Посилання Вікісховище: Bupleurum rotundifolium Віківиди: Bupleurum rotundifolium EOL: 581711 IPNI: 38361-2 ITIS: 29608 NCBI: 90446 Ласкавець круглолистий [1] [1] [2] ( Bupleurum rotundifolium ) — вид рослин родини Окружкові (Apiaceae)
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